In how many ways can 4 balls be placed in 8 boxes if the balls are indistinguishable, and the boxes are distinguishable?
THERE ARE 5 CASES
1ST CASE-1-1-1-1-0-0-0-0,TOTAL POSSIBLE =4*3*2*1*(8!/4!/4!(FOR BOX))=24*70=1680.
2ND CASE-2-1-1-0-0-0-0-0,TOTAL POSSIBLE =4C2*2*(8!/5!/2!)=2016.
3RD CASE-2-2-0-0-0-0-0-0,TOTAL POSSIBLE =4C2*2C2*8!/6!/2!=336.
4TH CASE-3-1-0-0-0-0-0-0,TOTAL POSSIBLE =4C3*1*8!/6!=224.
5TH CASE-4-0-0-0-0-0-0-0,TOTAL POSSIBLE=4C4*8!/7!=1*8=8.
Since the balls are indistinguishable, we can think about this problem as placing 4 identical labels into 8 distinct boxes. The first label can be placed in any of the 8 boxes, the second label can be placed in any of the remaining 7 boxes, and so on. So 8×7×6×5=1680.
The 8 boxes are all the same but the 4 balls arn't so the answer will be the same for 4,5,6 or any number more boxes.
The 4 balls can be in one box. 1 way
The are 4 ways to take one ball on it's own. so there are 4 ways to have a 3 to one split.
Say the balls are black, red, green and yellow.
To split them into two piles of 2 balls the black can pair with the red or the black leaving the green and yell ow together, OR
the black can pair with the green, OR the black can pair with the Yellow. 3 ways
So there is a total of 8 ways.